JEE Main 2023 — Mathematics Calculus

Given,

$${\int }_{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}....(1)$$∫0α​x+α​−x​x​dx=1516+202​​....(1)

Rationalise the denominator in L.H.S we get,

$${\int }_{0}^{\alpha }\frac{x(\sqrt{\alpha +x}+\sqrt{x})}{\alpha }dx=\frac{1}{\alpha }[{\int }_{0}^{\alpha }x\sqrt{\alpha +x}dx+{\int }_{0}^{\alpha }{x}^{3/2}dx]$$∫0α​αx(α+x​+x​)​dx=α1​[∫0α​xα+x​dx+∫0α​x3/2dx]

Put $\alpha +x={t}^{2}\Rightarrow dx=2\mathrm{tdt}$α+x=t2⇒dx=2tdt

Limit when $x=0\rightarrow t=\sqrt{\alpha },x=\alpha \rightarrow t=\sqrt{2\alpha }$x=0→t=α​,x=α→t=2α​

$$\frac{1}{\alpha }{\int }_{\sqrt{\alpha }}^{\sqrt{2\alpha }}({t}^{2}-\alpha )2{t}^{2}\mathrm{dt}+\frac{2{\alpha }^{3/2}}{5}$$α1​∫α​2α​​(t2−α)2t2dt+52α3/2​

$$=\frac{1}{\alpha }{[\frac{2{t}^{5}}{5}-\frac{2\alpha {t}^{3}}{3}]}_{\sqrt{\alpha }}^{\sqrt{2\alpha }}+\frac{2{\alpha }^{3/2}}{5}$$=α1​[52t5​−32αt3​]α​2α​​+52α3/2​

$$=\frac{4\alpha \sqrt{2\alpha }+10\alpha \sqrt{\alpha }}{15}$$=154α2α​+10αα​​

From equation $(1)$(1)

$${\int }_{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}$$∫0α​x+α​−x​x​dx=1516+202​​

$$\Rightarrow \frac{4\alpha \sqrt{2\alpha }+10\alpha \sqrt{\alpha }}{15}=\frac{16+20\sqrt{2}}{15}$$⇒154α2α​+10αα​​=1516+202​​

On comparing both sides we get,

$$\alpha =2$$α=2